If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin

|x|+||y|=10

put x=0

you get |y|=10 .....y=+-10

when you y=0

you get |x| = 10 ..... x=+-10

plot this on the co-ordinate plane.

you will get a rhombus
area of rhombus = 1/2 (d1 x d2)= 20X20/2= 200

First of all to simplify the given expression a little bit let's multiply it be 2: \(|\frac{x}{2}|+|\frac{y}{2}|=5\) --> \(|x|+|y|=10\).

Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0):
\(y=0\) --> \(|x|=10\) --> \(x=10\) and \(x=-10\);
\(x=0\) --> \(|y|=10\) --> \(y=10\) and \(y=-10\).

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):



You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

Answer: D.

Similar questions:
https://gmatclub.com/forum/m06-5-absolut ... 08191.html
https://gmatclub.com/forum/graphs-modulu ... 86549.html

Hope it's clear.

I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?

Given: \(|x|+|y|=20\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-x-y=10\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-x+y=10\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(x-y=10\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(x+y=10\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll get the figure shown in my previous post.

Hope it's clear.

Similar questions to practice:
the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html
in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html
if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html
what-is-the-area-of-the-region-enclosed-by-lines-y-x-x-y-150487.html
new-set-of-mixed-questions-150204-100.html#p1208441
the-area-bounded-by-the-three-curves-x-y-1-x-1-and-y-186595.html
if-equation-x-2-y-2-5-encloses-a-certain-region-126117.html

Hope this helps.

hey Bunuel,

I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.

We can re-write the question as below

\(x^2/4 +y^2/4 = 5\) (since \(|x| = x^2\))

\(x^2 + y^2 = 20\)

This is the equation is a circle having the centre at (0,0) (general form is \(x^2 + y^2= r^2\))

area =\(3.14 * R^2\) = \(3.14 * 20\) = 62.8

What am i assuming wrong here?? Thanks!

The part that I have highlighted above is WRONG which the first step in your solution

|x| is NOT equal to x^2 for all values of x[/highlight]

The Function "Modulus" only keeps the final sign Positive but that doesn't mean what you mentioned in the quoted Highlighted section.

Alternatively you can solve this question in this way

Step 1: Substitute y=0, \(|\frac{x}{2}| + |\frac{0}{2}| = 5\) i.e. \(|\frac{x}{2}| = 5\) i.e. \(|x| = 10\) i.e. \(x = +10\)

So on the X-Y plane you get two Point (+10,0) and (-10,0)

Step 2:Substitute x=0, \(|\frac{0}{2}| + |\frac{y}{2}| = 5\) i.e. \(|\frac{y}{2}| = 5\) i.e. \(|y| = 10\) i.e. \(y = +10\)

So on the X-Y plane you get two lines parallel to X-Axis passing through Y=+10 and Y=-10

So on the X-Y plane you get two Point (0, +10) and (0, -10)

Join all the four points, It's a Square with Side \(10\sqrt{2}\)

i.e. Area =\((10\sqrt{2})^2\) = 200

One More Clarification

( \(|x| is NOT equal to x^2\))

Instead, \(|x| = \sqrt{(x^2)}\)

why should I suppose that x or y equals +- 10 & zeros ? what about +- 5 as following :
|+ 5 |+ |-5| = 10
|-5|+|+5| = 10
|-5|+|-5|= 10
|+5|+|+5|=10
S0 we have Square with side of 10 length
Its area is 100

Hi Hatemnag,

The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane

One Linear equation when x is +ve and y is +ve i.e. X+Y = 10
Second Linear equation when x is +ve and y is -ve i.e. X-Y = 10
Third Linear equation when x is -ve and y is +ve i.e. -X+Y = 10
Forth Linear equation when x is -ve and y is -ve i.e. -X-Y = 10

So you need to plot these equation and then take the area of Quadrilateral formed

Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (-10,0), (0,10) and (0,-10)

Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the Square


I hope this clears your doubt!

Sorry, i dont know what i am missing, how do i get the diagonal to be 20?...from the square i got, i have all the sides equal to 20, hence the area=400.

Hi Jayanthjanardan,

The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane

One Linear equation when x is +ve and y is +ve i.e. X+Y = 10
Second Linear equation when x is +ve and y is -ve i.e. X-Y = 10
Third Linear equation when x is -ve and y is +ve i.e. -X+Y = 10
Forth Linear equation when x is -ve and y is -ve i.e. -X-Y = 10

NOTE: PLEASE PLOT THE LINES TO UNDERSTAND THE FIGURE (REFER THE FIGURE) and see that Diagonal of Square is 10

So you need to plot these equation and then take the area of Quadrilateral formed

Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (-10,0), (0,10) and (0,-10)

Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the Square


I hope this clears your doubt!

Hi GMATInight,

thanks a lot for the expalnantion. I get the logic now.

However, kindly refer to this link below.

in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549-40.html#p1540033

Its a similar problem, but the diagram we end getting is a square and not a rhombus...what am i missing here?

Even that is a square but never forget that a Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that fugure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won;t be wrong either but you are right about the figure being a Square.

|x| + |y| = 10.
Take the first quadrant => x + y = 10
Draw a line with (10,0) and (0,10) .. Area of this quadrant will be = 0.5 * 10 * 10 = 50
As this is mod it will be same in all quadrant = 50 * 4 = 200

Hi Bunuel and other experts,
Can someone help me draw the diagram because with diagram i am getting 40 not 20.

Here it is: